Superset of Neighborhood in Metric Space is Neighborhood

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $a \in A$ be a point in $M$.

Let $N$ be a neighborhood of $a$ in $M$.

Let $N \subseteq N' \subseteq A$.

Then $N'$ is a neighborhood of $a$ in $M$.

Proof

By definition of neighborhood:

$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq N$

where $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$ in $M$.

By Subset Relation is Transitive:

$\map {B_\epsilon} a \subseteq N'$

The result follows by definition of neighborhood of $a$.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8: \ N \, 3$